Question

Let $6x+4y+5z=4,x+5y+2z=12$ and $\frac{x-9}{2}=\frac{y+4}{-1}=\frac{z-5}{1}$ be two lines then

• A. the angle between them must be $\frac{\pi }{3}$
• B. the angle between them must be $co{s}^{-1}\frac{5}{6}$
• C. the plane containing them must be $x+y-z=0$
• D. they are non-coplanar

Answer 1

The correct option is A the angle between them must be $\frac{\pi }{3}$

The vector parallel to line of intersection of planes is
$\lambda \left|\begin{array}{ccc}i& j& k\\ 6& 4& -5\\ 1& -5& 2\end{array}\right|=-\lambda (17\hat{i}+17\hat{j}+34\hat{k})$
$=\lambda ‘(\hat{i}+\hat{j}+2\hat{k})(\lambda ‘isscalar)$
Now angle between the lines
$cos\theta =\frac{\lambda ‘(\hat{i}+\hat{j}+2\hat{k})\cdot (2\hat{i}-\hat{j}+\hat{k})}{\lambda ‘\sqrt{6}\times \sqrt{6}}=\frac{1}{2}$
$\Rightarrow \theta =\frac{\pi }{3}$